Whenever a photon is emitted by hydrogen in Balmerseries, it is followed by another photon in Lyman series. Whatwavelength does latter photon correspond to ?

The question states that whenever a photon is emitted by hydrogen in the Balmer series, it is followed by another photon in the Lyman series. We need to determine the wavelength of the latter photon.

To find the wavelength of the photon in the Lyman series, we can use the Rydberg formula. The Rydberg formula is given by:

1/λ = R * (1/n1^2 - 1/n2^2)

Where λ is the wavelength of the photon, R is the Rydberg constant, and n1 and n2 are the principal quantum numbers.

In the Balmer series, the principal quantum number n1 is 2. Since the question states that the photon in the Balmer series is followed by a photon in the Lyman series, we can assume that the principal quantum number n2 is 1.

Plugging these values into the Rydberg formula, we get:

1/λ = R * (1/2^2 - 1/1^2)

Simplifying the equation, we have:

1/λ = R * (1/4 - 1/1)

1/λ = R * (1/4 - 4/4)

1/λ = R * (-3/4)

To find the wavelength, we need to take the reciprocal of both sides of the equation:

λ = 4/(-3R)

Therefore, the wavelength of the photon in the Lyman series corresponds to 4/(-3R).